Question

x^2-4/x^2-2x

I need help finding the vertical and horizontal asymtptotes and what kind of symmetry if yall dont mind :D

Answer 1

Is the function :
\[f(x)=x^2-\frac4{x^2}-2x???\]

Answer 2

no its (x^2-4) divided by (x^2-2x)

Answer 3

LIke this :
\[f(x)=\frac{x^2-4}{x^2-2x}??\]

Answer 4

yes

Answer 5

First, you have to determine the domain of the function,
what is it ?

Answer 6

the domain is (negative infinity, 0)(0,2)(2,infinity)

Answer 7

Exactly :
The domaine is :
\[D=(-\infty, 0)\cup(0,2)\cup (2,+\infty)\]
So you have to calculate the limits at the bounds of D

Answer 8

yo lost me haha how do i do that

Answer 9

It is very simple, let's begin with the infinity :
\[\Large \lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}\frac{x^2-4}{x^2-2x}=\lim_{x\to+\infty}\frac{x^2}{x^2}=\lim_{x\to+\infty}1=1\]
and :
\[\Large \lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{x^2-4}{x^2-2x}=\lim_{x\to-\infty}\frac{x^2}{x^2}=\lim_{x\to-\infty}1=1\]
So we can say that the horizontal line with equation (y=1) is an horizontal asymptote in the neighborhood of +infty and -infty

Answer 10

i see okay what about the vertical asymptote

Answer 11

Now we calculate the limit at 0 from the two sides :
\[\Large \lim_{x\to0^-}f(x)=\frac{0^2-4}{0^+}=-\infty\]
and :
\[\Large \lim_{x\to0^+}f(x)=\frac{0^2-4}{0^-}=+\infty\]
So the line with equation (x=0) is a vertical asymptote.
the same thing with 2 :P

Answer 12

so the vertical asymptote is 0

Answer 13

A vertical asymptot is a line not a number O.o
After that: this function has 2 verticals asymptotes ;)

Answer 14

where would i plot the vertical asymptotes i mean. sorry!

Answer 15

at 0 and where else

Answer 16

and 2?

Answer 17

yes

Answer 18

okay one last thing! what about the symmetry how do i figure out if its symmetrical. it has something to do with f(-x)=f(x) but im not clear on it